Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either “M” or from 00 to 33 “X” followed by “S” or “L”. For example, sizes “M”, “XXS”, “L”, “XXXL” are valid and “XM”, “Z”, “XXXXL” are not.
There are nn winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can’t remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer nn (1≤n≤1001≤n≤100) — the number of T-shirts.
The ii-th of the next nn lines contains aiai — the size of the ii-th T-shirt of the list for the previous year.
The ii-th of the next nn lines contains bibi — the size of the ii-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list bb from the list aa.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace “M” with “S” and “S” in one of the occurrences of “XS” with “L”.
In the second example Ksenia should replace “L” in “XXXL” with “S”.
In the third example lists are equal.
题意：给你两组字符串，求把第一组字符串装换成第二组时要变化多少次。其中字符串形式是固定的几种，且变化时不能增多也不能减少。
解题思路：根据题意，当字符串长度为一时，有不同的情况。因此，可以把字符串为一的字符单独记录有多少个，然后比较即可。对于长度不是一的，可以每个不同的只需变化一次，排序后一次比较即可，也可以单独记录比较。
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
using namespace std;
int n;
vector<string>g[4],h[4];
int a1=0,a2=0,a3=0,b1=0,b2=0,b3=0;
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
string s;cin>>s;
if(s.size()==1){
if(s=="M")a1++;
if(s=="S")a2++;
if(s=="L")a3++;
continue;
}
g[s.size()-1].push_back(s);
}
for(int i=0;i<n;i++)
{
string s;cin>>s;
if(s.size()==1){
if(s=="M")b1++;
if(s=="S")b2++;
if(s=="L")b3++;
continue;
}
h[s.size()-1].push_back(s);
}
sort(g[1].begin(),g[1].end());
sort(g[2].begin(),g[2].end());
sort(g[3].begin(),g[3].end());
sort(h[1].begin(),h[1].end());
sort(h[2].begin(),h[2].end());
sort(h[3].begin(),h[3].end());
int ans=0;
ans=a1+a2+a3-min(a1,b1)-min(a2,b2)-min(a3,b3);
for(int i=0;i<g[1].size();i++)
{
if(g[1][i]!=h[1][i]){
ans++;
}
}
for(int i=0;i<g[2].size();i++)
{
if(g[2][i]!=h[2][i]){
ans++;
}
}
for(int i=0;i<g[3].size();i++)
{
if(g[3][i]!=h[3][i]){
ans++;
}
}
cout<<ans<<endl;
return 0;
}